Geometrical hint: Draw the sets $A, B$ and $C$. Identify what the left-hand side and right-hand side of the equation corresponds to on your drawing. Confirm that they are equal. Try to figure out why they are equal no matter how the sets intersect.
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ArthurJul 6 '13 at 12:47

if I draw it its ok but there is way to write it? not by drawing
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Ofir AttiaJul 6 '13 at 12:51

Axiomatic hint: Try to put into symbols what it means that $x$ is an element in the left-hand side and of the right-hand side of the equation. See if you can logically make each of them imply the other.
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ArthurJul 6 '13 at 12:51

5 Answers
5

For the first one, suppose that $(A \setminus B) \cup (B \setminus C)$ is not empty. Take any $x \in (A \setminus B) \cup (B \setminus C)$. Then either $x \in A \setminus B$ or $x \in B \setminus C$. Note that in this particular case, both cannot be true (why?). If $x \in A \setminus B$, then $x \in A$ and $x \not \in B$. If $x \in B \setminus C$, then $x \in B$ and $x \not \in C$. This does not imply that $x \in A \setminus C$. If $x \in A \setminus B$, one of the possibilities above, then this does not give us any information about whether $x \in C$.

The first one is not true if you don't require something extra. Consider the case where $A\cap B=\varnothing$, but neither $A$ nor $B$ is empty; and $C=\varnothing$.

For the second one, pick $x\in(A\setminus B)\setminus C$. Then $x\notin C$ and $x\in A\setminus B$, by the definition of the set. It follows that $x\notin C$ and $x\notin B$ and $x\in A$. Continue with the manipulation until you prove that $x\in A\setminus(B\cup C)$. Then prove the other inclusion in a similar fashion (by going in reverse, for example).